From Wikipedia
In a Markov process, one assumes that $i_1 < \cdots < i_n$. Then, because of the Markov property, $$ p_{i_1,\ldots,i_n}(f_1,\ldots,f_n)=p_{i_1}(f_1)p_{i_2;i_1}(f_2\mid f_1)\cdots p_{i_n;i_{n-1}}(f_n\mid f_{n-1}), $$ where the conditional probability $p_{i;j}(f_i\mid f_j)$ is the transition probability between the times $i>j$. So, the Chapman–Kolmogorov equation takes the form $$ p_{i_3;i_1}(f_3\mid f_1)=\int_{-\infty}^\infty p_{i_3;i_2}(f_3\mid f_2)p_{i_2;i_1}(f_2\mid f_1) \, df_2. $$
I was wondering if the reverse is true. I.e., if a process satisfies the above form of the Chapman–Kolmogorov equation, will it be Markovian? Thanks and regards!
